Question
In: Statistics and Probability
A simple random sample of 60 items resulted in a sample mean of 63. The population...
A simple random sample of 60 items resulted in a sample mean of 63. The population standard deviation is 12.
a. Compute the 95% confidence interval for the population mean (to 1 decimal).
( , )
b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean (to 2 decimals).
( , )
Solutions
Expert Solution
since population standard deviation is given so we use z-value to find confidence interval
(1-alpha)*100% confidence interval for population mean=sample mean±z(alpha/2)*sd/sqrt(n)
95% confidence interval for population mean=sample mean±z(0.05/2)*sd/sqrt(n)
(a) answer is (60,66)
95% confidence interval for population mean=sample mean±z(0.05/2)*sd/sqrt(n)=63±1.96*12/sqrt(60)=63±3.04=
=(59.96,66.04)=(60,66) ( one decimal place approximation)
| n= | 60 | |||
| sample mean= | 63 | |||
| s= | 12.00 | |||
| z-value | margin of error | lower limit | upper limit | |
| 95% confidence interval | 1.96 | 3.04 | 59.96 | 66.04 |
(b) answer is (60.85,65.15)
95% confidence interval for population mean=sample mean±z(0.05/2)*sd/sqrt(n)=63±1.96*12/sqrt(120)=63±2.15=
=(60.85,65.15) ( two decimal place approximation)
| n= | 120 | |||
| sample mean= | 63 | |||
| s= | 12.00 | |||
| z-value | margin of error | lower limit | upper limit | |
| 95% confidence interval | 1.96 | 2.15 | 60.85 | 65.15 |
orchestra answered 2 years agoRelated Solutions
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